find f(x) = x2 - 2x -2 find f' (x) using the long method
Let f(x)=x2−2x−2f(x) = x^2 - 2x -2f(x)=x2−2x−2. Let us find f′(x)f' (x)f′(x) using the long method.
f′(x)=limΔx→0f(x+Δx)−f(x)Δx=limΔx→0(x+Δx)2−2(x+Δx)−2−(x2−2x−2)Δx=limΔx→0x2+2xΔx+(Δx)2−2x−2Δx−2−x2+2x+2Δx=limΔx→02xΔx+(Δx)2−2ΔxΔx=limΔx→0(2x+Δx−2)=2x−2.f'(x)=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} \\=\lim\limits_{\Delta x\to 0}\frac{(x+\Delta x)^2 - 2(x+\Delta x) -2-(x^2 - 2x -2)}{\Delta x} \\=\lim\limits_{\Delta x\to 0}\frac{x^2+2x\Delta x+(\Delta x)^2 - 2x-2\Delta x -2-x^2 +2x +2}{\Delta x} \\=\lim\limits_{\Delta x\to 0}\frac{2x\Delta x+(\Delta x)^2 -2\Delta x }{\Delta x} \\=\lim\limits_{\Delta x\to 0}(2x+\Delta x -2 )\\=2x-2.f′(x)=Δx→0limΔxf(x+Δx)−f(x)=Δx→0limΔx(x+Δx)2−2(x+Δx)−2−(x2−2x−2)=Δx→0limΔxx2+2xΔx+(Δx)2−2x−2Δx−2−x2+2x+2=Δx→0limΔx2xΔx+(Δx)2−2Δx=Δx→0lim(2x+Δx−2)=2x−2.
